Calculation of $\bf{Max.}$ value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$

(1) Calculation of Max. and Min. value of $\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\;\; \forall x\in \mathbb{R}$

(2) Calculation of Max. value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$

My Try:: for $(1)$ one:: Using the Cauchy-Schwarz inequality

$$\Rightarrow \displaystyle \left\{\left(\sqrt{x^2-3x+2}\right)^2+\left(\sqrt{2+3x-x^2}\right)^2\right\}\cdot \left(1^2+1^2\right)\geq \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2$$ $$\Rightarrow \displaystyle \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2\leq 8$$

$$\Rightarrow \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}\leq 2\sqrt{2}$$

and equality hold when $\displaystyle \sqrt{x^2-3x+2}=\sqrt{2+3x-x^2}\Rightarrow x=0\;,3$

Is my solution right for Max.? If not then please help me, and how can I calculate for Min. and also help required in $(2)$ one.

• It's correct, you can check it here: wolframalpha.com/input/… – Zafer Cesur Jan 7 '14 at 14:49
• Too bad you cannot use derivatives. Your answer is correct. – user114628 Jan 7 '14 at 14:49
• Why can't we use derivatives BUT are allowed to assume knowledge of Cauchy-Schwarz inequality!? – Squirtle Jan 7 '14 at 15:06
• Since you classified this as "algebra-precalculus" (despite the mention of Cauchy-Schwartz inequality), I thought I'd try precalculus methods. Completing the square under the radicals in (1) leads to $\sqrt{u^2 + \frac{1}{4}} + \sqrt{-u^2 + \frac{17}{4}},$ where $u = x - \frac{3}{2}$ and $-\frac{1}{2}\sqrt{17} \leq u \leq \frac{1}{2}\sqrt{17}$ (the restriction on $u$ is from $-u^2 + \frac{17}{4} \geq 0$). At this point I don't see how to continue using only precalculus methods. (continued) – Dave L. Renfro Jan 7 '14 at 15:06
• (continuation) However, it's now an easy first semester "max/min on a closed and bounded interval" problem, with the max/min candidates winding up being $u = 0, \; \pm \sqrt{2}, \; \pm \frac{1}{2}\sqrt{17}.$ – Dave L. Renfro Jan 7 '14 at 15:07

(1):the approach is not right ,you can't make sure $a\ge b,b\le c \to a \ge c$
the easy way is let$u=x^2-3x, f(x)=f(u)=\sqrt{2+u}+\sqrt{2-u},f^2=4+2\sqrt{4-u^2}$
$0\le\sqrt{4-u^2} \le 2 \implies 4 \le f^2 \le 8 \implies 2 \le f \le 2\sqrt{2}$ when $u=0$ and $u=2$ get Max and Min.
(2): $\sqrt{10x-x^2}$ domain is [$0,10$], $\sqrt{18x-x^2-77}$ domain is [$7,11$] so $f(x)=\sqrt{10x-x^2}-\sqrt{18x-x^2-77}$ domain is [$7,10$]
$\sqrt{10x-x^2}$ is mono decreasing on [$7,10$], $\sqrt{18x-x^2-77}$ is increasing on [$7,9$] so we can sure $f(x)$ will get Max when $x=7$ when $x$ on [$7,9$], the remain is [$9,10$],but it is impossible for any $f(x) \ge f(7)$ because it is always $<f(7)- a, a >0$